The dP/dt of the adiabatic expansion is -42/11 kPa/min
<h3>How to calculate dP/dt in an adiabatic expansion?</h3>
An adiabatic process is a process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression
Given b=1.5, P=7 kPa, V=110 cm³, and dV/dt=40 cm³/min
PVᵇ = C
Taking logs of both sides gives:
ln P + b ln V = ln C
Taking partial derivatives gives:
Substitutituting the values b, P, V and dV/dt into the derivative above:
1/7 x dP/dt + 1.5/110 x 40 = 0
1/7 x dP/dt + 6/11 = 0
1/7 x dP/dt = - 6/11
dP/dt = - 6/11 x 7
dP/dt = -42/11 kPa/min
Therefore, the value of dP/dt is -42/11 kPa/min
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Answer:
5x^2-6x-10
Step-by-step explanation:
(2x^2-2x)-(-3x^2+4x+10)
2x^2-2x+3x^2-4x-10
5x^2-2x-4x-10
5x^2-6x-10
Answer:
The probability that <em>X</em> is less than 42 is 0.1271.
Step-by-step explanation:
The random variable <em>X </em>follows a Normal distribution.
The mean and standard deviation are:
E (X) = <em>μ</em> = 50.
SD (X) = <em>σ</em> = 7.
A normal distribution is continuous probability distribution.
The Normal probability distribution with mean µ and standard deviation σ is given by,
To compute the probability of a Normal random variable we first standardize the raw score.
The raw scores are standardized using the formula:
These standardized scores are known as <em>z</em>-scores and they follow normal distribution with mean 0 and standard deviation 1.
Compute the probability of (X < 42) as follows:
*Use a <em>z</em>-table for the probability.
Thus, the probability that <em>X</em> is less than 42 is 0.1271.
The normal curve is shown below.
Answer:
66 67/100
Step-by-step explanation:
Answer:
3
Step-by-step explanation:
Segment BC corresponds to segment DF. The length of BC is the distance between coordinates (0, 2) and (3, 2). These points are on the same horizontal line (y=2), so the distance between them is the difference of their x-coordinates: 3 - 0 = 3.
